# finding local extrema

I have a question that asks to find all local extrema for the following function $$f(x)=x^3-\sqrt{4x}$$

I started by finding computing the first derivative: $$f'(x)=\frac{3x^{5/2}-1}{\sqrt{x}}$$ $f'(x) = 0$ if $x=\left(\frac{1}{3}\right)^{2/5}$. Also, $f'$ is not defined when $x=0$. In finding the local extrema do I include $x=0$?

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Yes, you need to add $0$.

Study the variations of $f$ to convince yourself. It is nonincreasing on $[0,x]$ and nondecreasing on $[x,+\infty)$.

So $f(x)$ is actually a global minimum.

And since $\lim_{+\infty}f(x)=+\infty$, $f(0)$ is a local (not global) maximum.

Be careful, not every critical point is a local extremum. You need the first derivative test or the second derivative test, for instance, to conclude. That's why, anyhow, the best way is to fully study the variations of the function on its domain by determining the sign of its derivative, when it is available like here.

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Thanks for your response. So in terms of local extrema, $f(x)$ has a local minimum at $x=\left(\frac{1}{3}\right)^{2/5}$? – Paul Mar 6 '13 at 0:40
@Paul Yes. And it is even global. See also my edit about critical points. – 1015 Mar 6 '13 at 0:41